Let $I$ be a set, $C=(c_{ij})$ be a generalized Cartan matrix, $r$ be the rank of $C$, $I'$ be a subset of $I$ such that $(c_{ij}), i, j \in I'$ is invertable. Let $\mathfrak{g}$ be a Kac-Moody Lie algebra and $\mathfrak{h}$ be its Cartan subalgebra. $\mathfrak{g}$ is generated by $x_i^{\pm}$, $h_i$, $d_j$, $i,j\in I, j\not\in I'$. There is a unique linear isomorphism $\phi: \mathfrak{h}^{*} \mapsto \mathfrak{h}$, $\phi(\lambda)=t_{\lambda}$, where $t_{\lambda}$ is the unique element of $\mathfrak{h}$ satisfying $(t_{\lambda}, h)=\lambda(h)$, $(,)$ is a unique symmetric invariant bilinear form on $\mathfrak{g}$. Usually we define coroot of a root $\lambda$ by $\lambda^{\vee}=\frac{2\lambda}{(\lambda, \lambda)}$. But some authors define it differently. I am reading a survey paper about finite dimensional representations of quantum affine algebras (this survey does not have an electric version). In the paper, the author defines the following,
let $0 \neq \lambda \in \mathfrak{h}^*$, define $\lambda^{\vee}=\frac{2t_{\lambda}}{(\lambda, \lambda)}$. What are differences between this definition and the usually coroot? It is said that $\alpha_i^{\vee}=h_i$ and, if $\lambda=\sum_{i\in I} m_i\alpha_i\neq 0$, then $\lambda^{\vee}=\sum_{i\in I}\frac{s_i}{r^{\vee}}m_ih_i$. Here $r^{\vee}$ is the lacing number of $\mathfrak{g}$. Why these hold? Thank you.

1 Answer
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Note: I will consider the special case where $\mathfrak{g}$ is semisimple. Here, I understand the picture well. I believe that everything carries over to the Kac-Moody case without change, but I would like add this disclaimer, just in case there are some aspects of the theory that need modification.

Before we can talk about the differences in the definitions, we need to see where they come from. We are given an inner product on $\mathfrak{h}$ and an isomorphism $\mathfrak{h}\cong\mathfrak{h}^*$ induced by the inner product. We would like to define an inner product on $\mathfrak{h}^*$ so that the map is an isometry.

However, we have another relationship between $\mathfrak{h}$ and $\mathfrak{h}^*$, because for every root, $\alpha$, we have a $\mathfrak{sl}_2$ triple $(e_{\alpha},f_{\alpha},h_{\alpha})$, where $e_{\alpha}\in \mathfrak{g}_{\alpha}$, $f_{\alpha}\in \mathfrak{g}_{-\alpha}$, and $h_{\alpha}\in \mathfrak{h}$. While $e_{\alpha}$ and $f_{\alpha}$ are only well defined up to constants, $h_{\alpha}$ is unique (given the commutation relations). We study the representations of $\mathfrak{g}$ by restricting to these copies of $\mathfrak{sl}_2$ and using what we know of the representation theory of $\mathfrak{sl}_2$.

The question becomes, how does $\alpha$ relate to $\phi^{-1}(h_{\alpha})$? Quite simple, $\phi(h_{\alpha})=2\alpha/(\alpha,\alpha),$ and equivalently, $\phi(\alpha)=2h_{\alpha}/(h_,{\alpha},h_{\alpha})$. This comes from looking at properties of the Killing form and of $\mathfrak{sl}_2$.

So the question becomes, which is more natural: using $h_{\alpha}$, essential to the representation theory, or using $\phi^{-1}(h_{\alpha}$ so that much of what we describe can be done entirely in $\mathfrak{h}^*$, without mentioning $\mathfrak{h}$? My personal opinion is that for understanding, the former choice is superior, but for working and abstraction, the latter choice is. However, neither is great if you don't know where it comes from.